Measure Theory

Preface
Introduction
Sigma algebras
Bibliography

Preface

Basically, the people in the maths discord have announced a measure theory reading group that meets every Saturdays. I have been wanting to learn measure theory for a while and I see this as a perfect opportunity to finally start learning it given that it is enormously helpful having other people studying it together with you trying to cover a chapter each week. So, consider this blog as a place for my notes as I study measure theory. As a reference, the group will primarily be using Folland (1999) and so will I in these notes. If I use any other textbook, I will make sure to reference it later in this blog.

Introduction

We want a mapping $\mu: \mathscr{P}(\mathbb R^{n}) \to [0,\infty]$ that takes subsets of $\mathbb R^{n}$ , $n\in \mathbb N\backslash \{0\}$ to a non-negative real number (extended to $+\infty$ ). We call such a mapping the $n$ -dimensional measure of $E\subseteq{\mathbb R}$ given as $\mu(E)\in [0,\infty]$ . We want such a measure to satisfy the following few properties.

Properties:

P-1. If $E = \bigcup_{j\in \mathbb Z_{+}}E_{j}$ where $E_{i}\cap E_{j}=\phi$ for $i\neq j$ and each $E_{i}\subseteq \mathbb R^{n}$ then,

\mu(E) = \mu(\bigcup_{j\in\mathbb Z_{+}}E_{j}) = \sum_{j=1}^{\infty}\mu(E_{j})

P-2. If $E$ is congruent to $F$ , i.e., $E$ can be transformed to $F$ by translations, rotations, and reflections, then

\mu(E) = \mu(F)

P-3. The measure of the unit cube is 1.

\mu(Q) = 1

where $Q = \{x\in \mathbb R^{n} \mid 0\leq x_{j} < 1~\forall~j\in \{1,2,\dots, n\}\}$

It turns out that the three properties P-1, P-2 and P-3 are mutually inconsistent.

Theorem:

There is no such map

\mu: \mathscr{P}(\mathbb R) \to [0,\infty]

that satisfies all P-1, P-2 and P-3

Proof:

Define a relation $\sim$ on $[0,1)$ as $x\sim y \iff x-y\in \mathbb Q$ . Clearly, $\sim$ is an equivalence relation. Consider $\Lambda = % [0,1)\!/\!{\sim}%$ the set of all equivalence classes of $[0,1)$ under the equivalence relation $\sim$ .

Let us invoke the Axiom of Choice on $\Lambda$ to pick from each equivalence class in $\Lambda$ precisely one member and call this set $N$ such that $N\subseteq [0,1)$ .

Now, let $R = \mathbb Q \cap [0,1)$ . For each $r \in R$ , let

N_{r} = \{y+r\mid y\in N \cap [0,1-r)\} \cup \{y+r-1 \mid y\in N\cap [1-r, 1)\}.

$N_{r}$ is just every element of $N$ shifted to the right by rational $r$ units and the part that sticks out beyond $[0,1)$ is shifted back one unit to the left.

Then, each $N_{r}\subseteq [0,1)$ . Now, consider any $x\in [0,1)$ . Suppose $y\in N$ such that $y\in [x]$ , then $x\in N_{r}$ where $r = x-y$ if $x\geq y$ or $r = x-y+1$ if $x < y$ . If $x\in N_{r}\cap N_{s}$ then, $x-r$ or ( $x-r+1$ ) and $x-s$ or ( $x-s+1$ ) are two distinct elements of $N$ that are in the same equivalence class which is impossible.

From P-1 and P-2,

\mu(N) = \mu(N\cap [0,1-r)) + \mu(N\cap [1-r, 1)) = \mu(N_{r})

Since $[0,1)$ is the disjoint union of $N_{r}$ and $R$ is countable

\mu([0,1)) = \sum_{r\in R}\mu(N_{r})

From P-3, $\mu([0,1)) = 1$ and since $\mu(N_{r}) = \mu(N)$ , the sum on the right is either $0$ if $\mu(N) = 0$ or $\infty$ if $\mu(N)>0$ . So, the left and the right can never be equal. So, no such $\mu$ can exist.

So, what if we weaken the condition/property P-1? Let's say that the additivity holds only for finite sequences of subsets of $\mathbb R^{n}$ . This is a bad thing to do because for analysis, having countability is too important.

For dimensions $n\geq 3$ , there exists this thing called the Banach-Tarski Paradox whose result is that one can take a ball of the size of a pea, separate it to finite number of pieces and rearrange them to get a ball the size of the sun.

This prevents the construction of any map as we described above even for the restricted finite sequence additivity case that assigns postive, finite values for bounded open sets.

This means that $\mathbb R^{n}$ has subsets that can't be measured (no sensible measure can be defined for them). So, we will only define the measure for certain "nice" subsets of $\mathbb R^{n}$ .

Sigma algebras

Definition:

Let

X

be a nonempty set. An algebra of sets on

X

is a nonempty collection

\mathcal{A}

of subsets of

X

that is closed under finite unions and complements. That is,

E_{1}\dots E_{n} \in \mathcal{A} \implies \bigcup_{j}E_{j}\in \mathcal{A}

and

E\in \mathcal{A} \implies X\backslash E \in \mathcal{A}.

Definition:

\sigma

-algebra is an algebra that is closed under countable unions. It is also called a

\sigma

-field.

Proposition:

Algebras are closed under finite intersections.

Proof:

Let

\mathcal{A}

be an algebra on

X

. Consider

E_{1},\dots, E_{n} \in \mathcal{A}

. Then, for any

j\in \{1,\dots,n\},

E_{j} \in \mathcal{A} \implies X\backslash E_{j}\in \mathcal{A}

. Then,

\bigcup_{j}(X\backslash E_{j})\in \mathcal{A}

. This implies that

X\backslash \bigcup_{j}(X\backslash E_{j})\in \mathcal{A}.

By De-Morgan's law,

\bigcap_{j}E_{j} = X\backslash \bigcup_{j}(X\backslash E_{j}) \in \mathcal{A}

Proposition:

\sigma

-algebras are closed under countable intersections.

Proof:

Similar proof as the previous one can be applied as De-Morgan's laws apply for countable collections.

Proposition:

Let

X

be a nonempty set. Let

\mathcal{A}

be a

\sigma-algebra

X

. Then,

\phi\in \mathcal{A}

and

X \in \mathcal{A}

Proof:

Let

E\in \mathcal{A}

. This implies,

X\backslash E \in \mathcal{A} \implies [X = E\cup (X\backslash E) \in \mathcal{A}].

Also,

\phi = E\cap (X\backslash E)\in \mathcal{A}.

Definition:

A pair

(X,\mathcal{A})

consisting of a set

X

and a

\sigma

-algebra

\mathcal{A}

of subsets of

X

is called a measurable space.

Proposition:

If an algebra

\mathcal{A}

X

is closed under countable disjoint unions then it is a

\sigma

-algebra.

Proof:

$\mathcal{A}$ is an algebra so it is closed under complementation.

Suppose $\{E_{j}\}_{j=1}^{\infty}\subseteq \mathcal{A}$ . Set

F_{k} = E_{k} \backslash \bigcup_{j=1}^{n-1}E_{j} = E_{k} \cap (X\backslash \bigcup_{j=1}^{n-1}E_{j})

Clearly, $F_{k} \in \mathcal{A}$ for all $k\geq 1$ and the $F_{k}$ s are disjoint. Notice $\bigcup_{k=1}^{\infty}F_{k} = \bigcup_{j=1}^{\infty} E_{j}$ So, $\mathcal{A}$ is also closed under countable unions. Hence, $\mathcal{A}$ is a $\sigma$ -algebra.

Example:

If $X$ is any set, $\mathscr{P}(X)$ and $(X,\phi)$ are sigma algebras on $X$ .

The $\mathscr{P}(X)$ is set of all subsets of $X$ so it is clearly a sigma algebra.

Clearly ( $X,\phi$ ) is closed under complementation and union.

Theorem:

If $X$ is uncountable, then

\mathcal{A} = \{E\subseteq X \mid E \text{ is countable or } X\backslash E \text{ is countable }\}

is a $\sigma$ -algebra.

This $\sigma-algebra$ is called the $\sigma$ -algebra of countable or co-countable sets.

Proof:

Observe, $E\in \mathcal{A} \iff E \text{ countable or } X\backslash E \text{ countable }$ $\iff X\backslash E \text{ countable } \lor E \text{ countable }$ $\iff X\backslash E~countable~\lor X\backslash (X\backslash E)~countable~$ $\iff X\backslash E \in \mathcal{A}.$

So $\mathcal{A}$ is closed under complementation.

Spse $\{E_{j}\}_{j\geq 1}\subseteq \mathcal{A}$ .

If all $E_{j}$ are countable then clearly $\bigcup_{j\geq 1}E_{j}$ is countable and is in $\mathcal{A}$ .

Now, suppose for some $j\prime \geq 1, E_{j'}$ is co-countable.

Then notice

X\backslash \bigcup_{j\geq 1}E_{j} = \bigcap_{j\geq 1}(X\backslash E_{j}) \subseteq X\backslash E_{j\prime}

Since, $X\backslash E_{j\prime}$ is countable, $X\backslash \bigcup_{j\geq 1}E_{j}$ is countable and hence,

$\bigcup_{j\geq 1}E_{j}$ is co-countable.

Hence, $\mathcal{A}$ is closed under countable unions.

Definition:

Let

X

be a set and

\mathcal{F}\subseteq \mathscr{P}(X)

. The, the smallest $\sigma$ -algebra of subsets of

X

containing

\mathcal{F}

is called the $\sigma$ -algebra generated by $\mathcal{F}$ and is denoted by

\sigma(\mathcal{F}).

Proposition:

Let

X

be a set. For any family

\mathcal{F} \subseteq \mathscr{P}(X)

there exists a unique

\sigma

-algebra generated by

\mathcal{F}

Proof:

Existence of a $\sigma-$ algebra containing $\mathcal{F}$ follows from the existence of $\mathscr{P}(X)$ and an earlier example.

Now, set $\sigma(\mathcal{F}) = \bigcap\{A\mid (\mathcal{F}\subseteq A) \land A \text{ is a sigma algera on } X\}$

By construction $\mathcal{F}\subseteq \sigma(\mathcal{F})$ . Take any $\{A_{j}\}_{j\geq 1}\subseteq \sigma(\mathcal{F})$ . Then, each $A_{j}$ belongs to every $\sigma$ -algebra of $X$ that contains $\mathcal{F}$ . So, their complements and countable intersections, thereby countable unions belongs to any $\sigma$ -algebra of $X$ that contains $\mathcal{F}$ . Hence, $\sigma(\mathcal{F})$ is a $\sigma$ -algbera.

Uniqueness follows from the fact that existence of a $\sigma$ -algebra $\mathcal{B}$ that contains $\mathcal{F}$ but not $\sigma(\mathcal{F})$ contradicts the definition of $\sigma(\mathcal{F})$ since $\mathcal{B}\cap \sigma(\mathcal{F})$ contains $\mathcal{F}$ and is a $\sigma$ -algebra.

Notice that the family of intervals with rational endpoints is countable. However, also notice that this family generates a $\sigma$ -algebra that contains all single-point sets.

Why? well the Borel $\sigma$ -algebra contains all singletons and every open set can be written as a countable union of intervals with rational endpoints and similar arguments one can show this.

This means that a countable family of subsets of $X$ can generate a $\sigma$ -algebra that is uncountable.

Lemma:

\mathcal{E}\subseteq \sigma(\mathcal{F})

then

\sigma(\mathcal{E})\subseteq \sigma(\mathcal{F})

Since, $\sigma(\mathcal{F})$ is a $\sigma$ -algebra containing $\mathcal{E}$ ; it must contain $\sigma(\mathcal{E})$ .

The $\sigma$ -algebra generated by the class $\mathcal{F}$ can not be described in a constructive form by means of countable unions, intersections or complements of elements of $\mathcal{F}$ . This is because these operations can be repeated unlimited number of times and we can obtain new classes all the time, but their union does not exhaust the $\sigma$ -algebra generated by $\mathcal{F}$ .

The following is an example where one can explicitly describe the $\sigma$ -algebra generated by a class of sets.

Example:

Let $\mathcal{A}_{0}$ be a $\sigma$ -algebra on a space $X$ . Suppose $S\subseteq X$ and $S\notin \mathcal{A}_{0}$ . Then, the $\sigma$ -algebra $\sigma(\mathcal{A}_{0}\cup \{S\})$ , generated by $\mathcal{A}_{0}$ and the set $S$ coincides with the collection of all sets of the form

E = (A\cap S) \cup (B\cap (X\backslash S))

Proof:

Consider

\mathcal{F} = \{E \mid E = (A\cap S) \cup (B\cap X\backslash S)\, A,B\in \mathcal{A}_{0}\}

Bibliography

G. B. Folland, Real analysis: modern techniques and their applications, 2nd ed. New York: Wiley, 1999.